Cells, Emf, Internal Resistance

The cell, and what emf means

• A cell has two electrodes, the positive P and the negative N, immersed in an electrolytic solution. Dipped in the solution, the electrodes exchange charges with the electrolyte.
• The positive electrode acquires a potential difference $V_+ > 0$ with respect to the electrolyte immediately next to it (point A), and the negative electrode a potential $-V_-$ with respect to the electrolyte next to it (point B), with $V_- \ge 0$.
• When no current flows, the electrolyte is all at one potential, so the potential difference between P and N is $V_+ - (-V_-) = V_+ + V_-$. This is the electromotive force (emf) of the cell:

$$\varepsilon = V_+ + V_- > 0$$

• Emf is a potential difference, not a force; the name survives from a time when the phenomenon was not understood. Equivalently: the emf of a source is the work done per unit charge by the source in carrying charge from its lower-potential terminal to its higher-potential one.
• Definition to write for a mark: emf is the potential difference between the terminals of a cell in an open circuit, i.e. when no current is being drawn from it. Unit: volt.

Internal resistance and the terminal voltage

• The electrolyte through which the current flows has a finite resistance $r$, called the internal resistance of the cell.
• With $R$ infinite, no current flows, and going from P through A and B to N gives $V = \varepsilon$: the emf is the open-circuit terminal potential difference.
• With $R$ finite a current $I$ flows, and the potential difference between P and N becomes

$$V = V_+ + V_- - Ir = \varepsilon - Ir$$

• The minus sign appears because inside the cell the current flows from B to A, i.e. against the fall of potential. The terminal voltage is therefore always less than the emf while the cell is discharging.
• Internal resistances may be neglected in practical calculations whenever $\varepsilon \gg Ir$. Dry cells have much higher internal resistances than ordinary electrolytic cells.
• When the cell is being charged, the external source drives current into the positive terminal, the sign of $Ir$ reverses, and

$$V = \varepsilon + Ir > \varepsilon$$

The circuit relation

• $V$ is also the potential difference across the external resistance, so $V = IR$. Combining with $V = \varepsilon - Ir$:

$$IR = \varepsilon - Ir \qquad \Rightarrow \qquad I = \dfrac{\varepsilon}{R + r}, \qquad V = \dfrac{\varepsilon R}{R + r}$$

• The maximum current is drawn when $R = 0$: $I_{\max} = \varepsilon / r$. In most cells the maximum permitted current is far lower than this, to avoid permanent damage.
• Internal resistance from measurements: $r = (\varepsilon - V)/I$, or in terms of the external resistance, $r = R(\varepsilon - V)/V$.
• Inside the cell the current flows from the negative terminal to the positive terminal; outside, from the positive terminal to the negative terminal through the external circuit.

Emf against terminal voltage — the 2-mark table

The two graphs, and how to get $\varepsilon$ and $r$ from them

• $V$ against $I$. Write $V = \varepsilon - Ir$ and compare with $y = mx + c$: it is a straight line with intercept $c = \varepsilon$ on the $V$-axis and slope $m = -r$. It meets the $I$-axis at $I = \varepsilon/r$.
• Hence the $V$-intercept of the line gives the emf of the cell, and the magnitude of its slope gives the internal resistance. That is the whole answer to "how can $\varepsilon$ and $r$ be determined from the plot".
• Worked reading: a line falling from $V = 5.6\ \text{V}$ at $I = 0$ through $4.2\ \text{V}$ at $I = 0.5\ \text{A}$ has slope $(4.2-5.6)/0.5 = -2.8$, so $r = 2.8\ \Omega$ and $\varepsilon = 5.6\ \text{V}$.
• $V$ against $R$. Write $V = \varepsilon R/(R+r)$: the curve starts at $V = 0$ when $R = 0$, rises steeply at first, then flattens and approaches $V = \varepsilon$ asymptotically as $R$ becomes very large. It never actually reaches $\varepsilon$.
• The current $I = \varepsilon/(R+r)$ is largest at $R = 0$ and falls monotonically as $R$ increases.

Worked cases CBSE has set

• Emf and $r$ from two readings. $R = 4\ \Omega$ gives $I = 1\ \text{A}$, so $\varepsilon = r + 4$. $R = 9\ \Omega$ gives $I = 0.5\ \text{A}$, so $r + 9 = 2\varepsilon = 2(r+4)$, giving $r = 1\ \Omega$ and $\varepsilon = 5\ \text{V}$.
• Ammeter and voltmeter readings. A 12 V source of internal resistance $2\ \Omega$ across $4\ \Omega$: $I = 12/(4+2) = 2\ \text{A}$ (ammeter), and the terminal voltage $= 12 - 2\times2 = 8\ \text{V}$ (voltmeter).
• Charging a battery. An 8 V battery of internal resistance $0.5\ \Omega$ charged from a 120 V dc supply through a series resistor of $15.5\ \Omega$: the net driving emf is $120 - 8 = 112\ \text{V}$, the total resistance $15.5 + 0.5 = 16\ \Omega$, so $I = 112/16 = 7\ \text{A}$, and the terminal voltage of the battery is $8 + 7\times0.5 = 11.5\ \text{V}$. The series resistor is there to limit the charging current to a safe value — without it the battery's very low internal resistance would let a dangerously large current flow.
• Drawing a large current safely. Since $I = \varepsilon/(R+r)$, a low-voltage battery must have a very low internal resistance to deliver a large current, whereas a high-voltage battery can afford a comparatively larger internal resistance and still supply the same current.

Standard results

• $\varepsilon = V_+ + V_-$; $V = \varepsilon - Ir$ (discharging), $V = \varepsilon + Ir$ (charging).
• $I = \varepsilon/(R+r)$, $V = \varepsilon R/(R+r)$, $I_{\max} = \varepsilon/r$ at $R = 0$.
• $r = (\varepsilon - V)/I = R(\varepsilon - V)/V$.
• $V$ against $I$: straight line, intercept $\varepsilon$, slope $-r$. $V$ against $R$: rising curve, asymptote $V = \varepsilon$.
• Current runs N to P inside the cell, P to N outside it.